Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{3t - 24}{-3t + 6} \div \dfrac{4t^2 - 32t - 36}{t^2 - t - 2} $
Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{3t - 24}{-3t + 6} \times \dfrac{t^2 - t - 2}{4t^2 - 32t - 36} $ First factor out any common factors. $y = \dfrac{3(t - 8)}{-3(t - 2)} \times \dfrac{t^2 - t - 2}{4(t^2 - 8t - 9)} $ Then factor the quadratic expressions. $y = \dfrac {3(t - 8)} {-3(t - 2)} \times \dfrac {(t + 1)(t - 2)} {4(t + 1)(t - 9)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {3(t - 8) \times (t + 1)(t - 2) } {-3(t - 2) \times 4(t + 1)(t - 9) } $ $y = \dfrac {3(t + 1)(t - 2)(t - 8)} {-12(t + 1)(t - 9)(t - 2)} $ Notice that $(t + 1)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {3\cancel{(t + 1)}(t - 2)(t - 8)} {-12\cancel{(t + 1)}(t - 9)(t - 2)} $ We are dividing by $t + 1$ , so $t + 1 \neq 0$ Therefore, $t \neq -1$ $y = \dfrac {3\cancel{(t + 1)}\cancel{(t - 2)}(t - 8)} {-12\cancel{(t + 1)}(t - 9)\cancel{(t - 2)}} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $y = \dfrac {3(t - 8)} {-12(t - 9)} $ $ y = \dfrac{-(t - 8)}{4(t - 9)}; t \neq -1; t \neq 2 $